# Under what condition the covariant derivative of Ricci operator along Killing vector field vanish?

Let $(M,g)$ be a Riemannian manifold and $V$ a unit Killing vector field on it. Under what condition on curvature tensor the following equation hold: $$\nabla_VQ=0,$$ where $Q$ is the Ricci operator defined as $g(QX,Y)=\rho(X,Y)$.

Update1: Einstein metrics that admit a unit Killing vector field satisfies the above equation.

Update2: Can anybody give an example other than Einstein manifolds that satisfy the above equation?

• Cartesian product of a sphere with a hyperbolic manifold. – Deane Yang Aug 3 '17 at 13:35

If $V$ is a Killing vector, then necessarily its Lie derivative annihilates all curvature tensors, including the Ricci operator, namely $$\mathcal{L}_V Q = \nabla_V Q + Q J_V - J_V Q = 0 ,$$ where $J_V$ is the ("Jacobian") endomorphism on tangent vectors defined by $\nabla_U V = J_V U$. Thus, the condition that you want seems to be just $Q J_V - J_V Q = 0$. I don't see how the unit condition on $V$ enters.
• @C.F.G I'm not sure I understand your question. A condition on the Ricci operator is by definition also a condition on the Riemann tensor. Maybe you are looking for a condition independent of $V$? – Igor Khavkine Aug 3 '17 at 4:12
• You are right. I want to know if we assume that $(M,g)$ is Einstein manifold then can be determine curvature tensor completely? Answer of my second update is positive or not? – C.F.G Aug 3 '17 at 6:05